lượng giác 11

Question

$cosx -\sqrt{3}sinx = 2cos3x$

$ sin9x +\sqrt{3}cos7x = sin7x +\sqrt{3}cos9x$

BMĐ · Answer 1 · 2/17/2013 10:40:50 AM

2.
Phương trình đã cho tương đương với:
$\sin9x-\sqrt3\cos9x=\sin7x-\sqrt3\cos7x$
$\Leftrightarrow \dfrac{1}{2}\sin9x-\dfrac{\sqrt3}{2}\cos9x=\dfrac{1}{2}\sin7x-\dfrac{\sqrt3}{2}\cos7x$
$\Leftrightarrow \sin(9x-\dfrac{\pi}{3})=\sin(7x-\dfrac{\pi}{3})$
$\Leftrightarrow \left[\begin{array}{l}9x-\dfrac{\pi}{3}=7x-\dfrac{\pi}{3}+k2\pi\\9x-\dfrac{\pi}{3}=\dfrac{4\pi}{3}-7x+k2\pi\end{array}\right.,k\in\mathbb{Z}$
$\Leftrightarrow \left[\begin{array}{l}x=k\pi\\x=\dfrac{5\pi}{48}+k\dfrac{\pi}{8}\end{array}\right.,k\in\mathbb{Z}$

lịch sử

Sửa 17-02-13 10:40 AM

BMĐ
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score 3 · Answer 2

Bình chọn tăng 3 Bình chọn giảm

1.
Phương trình đã cho tương đương với:
$\cos x-\sqrt3\sin x=2\cos3x$
$\Leftrightarrow \dfrac{1}{2}\cos x-\dfrac{\sqrt3}{2}\sin x=\cos3x$
$\Leftrightarrow \cos(x+\dfrac{\pi}{3})=\cos3x$
$\Leftrightarrow \left[\begin{array}{l}x+\dfrac{\pi}{3}=3x+k2\pi\\x+\dfrac{\pi}{3}=-3x+k2\pi\end{array}\right.,k\in\mathbb{Z}$
$\Leftrightarrow \left[\begin{array}{l}x=\dfrac{\pi}{6}-k\pi\\x=\dfrac{-\pi}{12}+k\dfrac{\pi}{2}\end{array}\right.,k\in\mathbb{Z}$